64
$\begingroup$

If I know $\|f_{n}(x) - f(x)\|_{L^{p}(\mathbb{R})} \rightarrow 0$ as $n \rightarrow \infty$, do I know $\lim_{n \rightarrow \infty}f_{n}(x) = f(x)$ for almost every $x$?

$\endgroup$
  • 4
    $\begingroup$ For sure along a subsequence. This is the first step when you prove that $L^p$ is a complete normed space. See johndcook.com/modes_of_convergence.html $\endgroup$ – Siminore Apr 28 '12 at 14:10
  • 1
    $\begingroup$ Choosing suitable intervals with length going to zero you can exhibit a sequence of L^p, but pointwise the sequence doesn't converges at any point! $\endgroup$ – checkmath Apr 28 '12 at 14:25
  • 29
    $\begingroup$ Look at $\chi_{[0,1]}$, $\chi_{[0,1/2]}$, $\chi_{[1/2,1]}$, $\chi_{[0,1/4]}$, $\chi_{[1/4,1/2]}$, $\ldots\,$. $\endgroup$ – David Mitra Apr 28 '12 at 14:36
  • 1
    $\begingroup$ If we add a condition: $(f_n)_{n=1}^\infty$ is a sequence of nonnegative functions and $f_{n+1}(x)\geq f(x)$, then will $L^p$-norm convergence imply pointwise convergence? $\endgroup$ – Shiquan May 26 '14 at 13:36
  • $\begingroup$ If we repalce $\mathbb R$ by a compact region, and assume that $f_n$ are smooth, does the statement hold? $\endgroup$ – DLIN Mar 21 '18 at 13:57
44
$\begingroup$

Let $\rho(x)=\chi_{[0,1]}$ be the characteristic function of the interval $[0,1]$. Then take the "dancing" sequence

$$ f_n(x) = \rho(2^mx-k) $$

where $n=2^m+k$ with $0\leq k<2^m$. This sequence converges to $0$ in $L^p$ but for any $x\in(0,1)$ we have $f_n(x)$ is not convergent.

However, it is a general fact that one can always extract a subsequence converging almost everywhere to $f$.

$\endgroup$
  • $\begingroup$ This is also sometimes called the typewriter sequence. $\endgroup$ – Watson Dec 29 '16 at 12:09
17
$\begingroup$

For $n$ integer and $0\leq k\leq 2^n-1$, denote $f_{n,k}:=n\chi_{[k2^{-n},(k+1)2^{-n}]}$. Consider $f_{1,0},f_{1,1},f_{2,0},f_{2,1},f_{2,2},f_{2,3},\ldots$ (put $g_n:=f_{\alpha_n,n-\alpha_n}$ where $\alpha$ is an integer such that $1+\ldots+2^{\alpha_n}\leq n<1+\ldots+2^{\alpha_n+1}$). Then $g_n$ doesn't converge to $0$ for any $x$, but converge to $0$ in $L^p$ for $1\leq p<\infty$. However, we can find an almost everywhere converging subsequence (here we can pick $g_{1+\ldots+2^n}$).

More generally, if $f_n\to 0$ in $L^p$, with $1\leq p<\infty$, we get can find $n_k$ increasing to $+\infty$ such that $\lVert f_{n_k}\rVert_{L^p}\leq 2^{-k}$. Let $g_k:=|f_{n_k}|$. Then $\mu\{x:g_k(x)\geq n^{-1}\}\leq \frac {2^{-kp}}{n^{-p}}$, so by Borel-Cantelli $\mu\{\limsup_k g_k\geq \frac 1n\}=0$ for each $n$ and $g_k\to 0$ almost everywhere.

$\endgroup$
15
$\begingroup$

Consider the "typewriter sequence" defined by the formula: $$f_n:=1_{\left[\frac{n-2^k}{2^k},\frac{n-2^k+1}{2^k}\right]}$$ where $k$ is an integer such that $2^k\leq n<2^{k+1}$, the sequence converges to zero in $L^p$ norm, but not pointwise.

$\endgroup$
  • $\begingroup$ Does $p=\infty$ also counts here? $\endgroup$ – Error 404 Oct 6 '18 at 16:28
  • $\begingroup$ @Error404 Of course, $p=\infty$ is not included here. Since converging in $L^\infty$-norm means converging uniformly a.e.; hence pointwise a.e. $\endgroup$ – Xiang Yu Oct 10 '18 at 14:22
  • $\begingroup$ Thanks for your reply and nice answer! $\endgroup$ – Error 404 Oct 10 '18 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.